\newproblem{lay:6_5_19}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.5.19}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ be an $m\times n$ matrix. Use the steps below to show that a vector $\mathbf{x}\in\mathbb{R}^n$ satisfies $A\mathbf{x}=\mathbf{0}$ if and only if
	$A^TA\mathbf{x}=\mathbf{0}$. This will show that $\mathrm{Nul}\{A\}=\mathrm{Nul}\{A^TA\}$.
	\begin{enumerate}[a.]
		\item Show that if $A\mathbf{x}=\mathbf{0}$, then $A^TA\mathbf{x}=\mathbf{0}$
		\item Suppose $A^TA\mathbf{x}=\mathbf{0}$. Explain why $\mathbf{x}^TA^TA\mathbf{x}=0$, and use this to show that $A\mathbf{x}=\mathbf{0}$
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}[a.]
		\item Let us assume that
		      \begin{center}
						$A\mathbf{x}=\mathbf{0}$
					\end{center}
					Multiplying on the left by $A^T$, we get
		      \begin{center}
						$A^TA\mathbf{x}=A^T\mathbf{0}=\mathbf{0}$
					\end{center}
		\item Let us assume that 
		      \begin{center}
						$A^TA\mathbf{x}=\mathbf{0}$
					\end{center}
					Multiplying both sides by $\mathbf{x}^T$, we get
		      \begin{center}
						$\mathbf{x}^TA^TA\mathbf{x}=\mathbf{x}^T\mathbf{0}=0$
					\end{center}
					But this means that the norm of $A\mathbf{x}$ is null because
		      \begin{center}
						$\mathbf{x}^TA^TA\mathbf{x}=\|A\mathbf{x}\|^2=0$
					\end{center}
					So $A\mathbf{x}=\mathbf{0}$.
	\end{enumerate}
}
\useproblem{lay:6_5_19}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
